A multivariable chain rule problem

Let [tex]f:\mathbb{R}^3 \rightarrow \mathbb{R}[/tex] and [tex]g:\mathbb{R}^2\rightarrow \mathbb{R}[/tex] be differentiable. Let [tex]F:\mathbb{R}^2 \rightarrow \mathbb{R}[/tex] be defined by the equation
[tex]F(x,y)=f(x,y,g(x,y)).[/tex]
Find [tex]DF[/tex] in terms of the partials of [tex]f[/tex] and [tex]g[/tex].

I think I see where you are getting confused. Hint: There is no such thing as [tex]D_3F[/tex] :-o

Since F is a function of [tex]\mathbb{R}^2[/tex], [tex]D_1F = \frac{dF}{dx}[/tex], and [tex]D_2F = \frac{dF}{dy}[/tex]. Therefore, the gradient (DF) is going to be [tex](D_1F, D_2F)[/tex].

The notation is misleading. You probably want to write out the definition of the partial derivative, [tex]D_1 F = \frac{F(x+\epsilon, y) - F(x, y)}{\epsilon}[/tex], (and similarly for [tex]D_2 F[/tex]). Only then should you expand F out.